1 / 23

A Quadratic Lower Bound for Three-Query Linear Locally Decodable Codes Over Any Field

A Quadratic Lower Bound for Three-Query Linear Locally Decodable Codes Over Any Field. David Woodruff IBM Almaden. Linear Locally Decodable Codes.

tbahena
Download Presentation

A Quadratic Lower Bound for Three-Query Linear Locally Decodable Codes Over Any Field

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. A Quadratic Lower Bound for Three-Query Linear Locally Decodable Codes Over Any Field David Woodruff IBM Almaden

  2. Linear Locally Decodable Codes A (q, , )-linearlocally decodable code (LDC) C: Fn! Fm is a linear map with a (possibly adaptive) decoder A such that 8x 2 Fn, 8 i 2 [n] and 8 y for which (y, C(x)) < m, 1. Pr [A(i, y) = xi] ¸ 1/|F| +  2. A queries at most q positions of y

  3. Applications • LDCs • Local Decoding of Large Files • Private Information Retrieval • Complexity Theory • Linearity important for various applications • Succinct representation / efficient encoding • Streaming • Matrix rigidity • … • Infinite fields, e.g., F = R, important in these applications

  4. Previous Bounds • Assume ,  are constants • q = 1: LDCs do not exist [KT] • q = 2:m = exp(n) for any field F [DS, KdW, Hadamard] • q = 3:for general |F|, only trivial m = (n) Upper bounds have m super-polynomial in n, but sub-exponential [Y, E]

  5. Our Result For any constants , , and any field F, we show for q = 3, any linear LDC C: Fn! Fm satisfies m = (n2) First non-trivial lower-bound for general F -Result holds, e.g., if F is the field of real or complex numbers

  6. Proof Overview 1. Make the decoder non-adaptive 2. Find a special set in the recovery graphs 3. The projection 4. Recursive projection

  7. Making the Decoder Non-Adaptive Lemma: Any (3, , )-LDC can be made into a (3, ’, ’)-LDC with a non-adaptive decoder -If,  are constant, then so are ’, ’ - Works for any field F Combining this with previous work, seems possible to get - (n2 / (|F| log2 n)) bound for q = 3 [KdW] - (n3/2) bound for q = 3 for any F [KT] In contrast, we will get (n2) for any F

  8. Proof of Lemma Lemma: Any (3, , )-LDC can be made into a (3, ’, ’)-LDC with a non-adaptive decoder -If,  are constant, then so are ’, ’ - Works for any field F Proof: Since LDC is linear, there are vectors v1, …, vm2 Fn for which for all x: C(x) = <v1, x>, <v2, x>, …, <vm, x> For each standard unit vector ei, there must be a matching Mi of (m) disjoint triples {vi1, vi2, vi3} which span ei The new decoder chooses such a triple uniformly at random

  9. Proof Overview 1. Make the decoder non-adaptive 2. Find a special set in the recovery graphs 3. The projection 4. Recursive projection

  10. The Recovery Hypergraphs • Vertices of G are v1, …, vm • Hyperedges of G are 3-edges which occur in some Mi • the 3-edge is then labeled ei • (mn) hyperedges • Lemma: there is a non-empty sub-hypergraph G’ µ G with minimum degree ¯n for a constant ¯ > 0 • Proof: iteratively remove minimum degree vertex until minimum degree larger than original average degree / 3

  11. Finding a Special Set • Choose any v 2 G’, and consider N(v), its set of neighbors in G’ • Since v has degree >¯n and occurs at most once in each Mi, from {v} [ N(v), we can span (n) different ei • Hence, |N(v)| = (n)

  12. The Picture v … e3 en e1 N(v) … e9 e2 e1 e101 N(N(v)) We know m ¸ |N(N(v))|, so let’s lower bound |N(N(v))|

  13. Proof Overview 1. Make the decoder non-adaptive 2. Find a special set in the recovery graphs 3. The projection 4. Recursive projection

  14. Setup • We have a minimum degree ¯n hypergraph G’ whose hyperedges are sets in Mi • We have found a set N(v) with |N(v)| > ¯ n • By definition of G’,N(v) is incident to (n2) hyperedges. • Let S ½ N(v) be an arbitrary subset of ¯ n/2 linearly independent vertices • Let E be a set of n-¯ n/2 standard unit vectors for which E [ S is a basis of Fn

  15. The Projection • By definition of G’,N(v) is incident to (n2) hyperedges. • Let S ½ N(v) be an arbitrary subset of ¯ n/2 linearly independent vertices • Let E be a set of n-¯ n/2 standard unit vectors for which E [ S is a basis of Fn Define a linear map L: L(s) = 0 for all s 2 S L(e) = e for all e 2 E

  16. The New Picture v … e3 en e1 0 a5 a6 S a3 a4 0 a1 a2 0 0 0 0 … 3-edges preserved by L e9 e2 e1 e101 N(S) c3 c4 c5 c6 b7 c7 c2 b3 b4 b5 b6 b1 c1 b2 Apply linear map L

  17. Reduction to Two Queries 0 0 0 0 0 0 … e9 e2 e1 e101 c3 c4 c5 c6 c7 c1 c2 • Each vertex in S has degree > ¯n, so at least ¯ n/2 • of 3-edges incident to it are preserved by L • We get (n2)2-edges on a set of |N(S)| vertices

  18. Isoperimetric Inequality • [Bollobas, GKST, DS]: Given r vectors in Fn for which for each i 2 [n], there is a matching Mi’ of 2-edges for which for each {a, b} 2 Mi’, ei2 span(a, b), then: • r log r = (Σi=1n |Mi’|) • In our setting, • r = |N(S)| • Σi=1n |Mi’| = (n2) • Hence, m ¸ |N(S)| = (n2/log n)

  19. Proof Overview 1. Make the decoder non-adaptive 2. Find a special set in the recovery graphs 3. The projection 4. Recursive projection

  20. Boosting the Lower Bound • If |N(S)| = (n2), then done. So suppose |N(S)| = o(n2) • Let’s also project a random constant fraction of eito 0 0 … e3 en e1 0 0 0 0 0 0 connected components … e9 e2 e1 e101 • Each component has rank at most 1

  21. The Connected Components • We get a collection of rank 1 components: … Can’t have a bunch of components with a constant number of vertices … C1 C2 Cs • With good probability,(n2) edges from S to N(S) have their labels projected to 0 • Let cibe the # of vertices in Ci, let e(Ci) be the # of edges • Σi=1s ci· |N(S)| = o(n2) • Σi=1s e(Ci) = (n2) • But, each Ci obeys isoperimetric inequality! • e(Ci) · ci log ci Projecting one vertex in each of the n/100 largest components to zero, we project a much larger set of vertices to 0

  22. Wrapping Up • To get (n2), repeatedly project large connected components to 0, then a new fraction of standard unit vectors adjacent to these components, obtaining new components, etc. • Gradually enlarge the set of vertices that is projected to 0 while preserving a large fraction of standard unit vectors • Summary: we show for any constants , , and any field F, any 3-query linear LDC C: Fn! Fm satisfies m = (n2)

  23. An (n2/log log n) Lower Bound Suppose |N(S)| = o(n2/log log n): • Σi=1s ci = o(n2/ log log n) • Σi=1s e(Ci) = (n2), and soΣi=1s ci log ci = (n2) Order the Ci so that c1¸ c2¸ … ¸ cs Lemma:Σi=1t ci = (n log n / log log n) for t = n/100 Proof: If not, then cj for j ¸ t, is o(log n / log log n). But then: (n2) = Σi=1s e(Ci) · o(n log2 n / log log n) + Σj=(s+1)t cj log cj But Σj=(s+1)t cj log cj = o(Σi=1s ci log log n), a contradiction. If we project one vertex in each of C1, …, Ct to 0, we project a set of (n log n / log log n) vertices to 0. New set incident to (n2 log n / log log n)3-edges, and we can lower bound neighborhood

More Related